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A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. Its...

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yakar | Student | eNoter

Posted April 26, 2012 at 2:17 PM via web

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A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. Its maximum cruising speed is 120mi/h. Calculate for distance and speed.

A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. Its maximum cruising speed is 120mi/h. (Round your answers to three decimal places.)

(a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?
1     ______________mi

(b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions?
2     _______________mi

(c) Find the minimum time that the train takes to travel between two consecutive stations that are 60 miles apart.
3     ____________min

(d) The trip from one station to the next takes at minimum 37.5 minutes. How far apart are the stations?
4     _____________mi

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thilina-g | College Teacher | (Level 1) Educator

Posted April 26, 2012 at 5:07 PM (Answer #1)

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1 mile = 5280 feet

We can first convert the acceleration value to mi/h2 and that would be more convenient.

10 ft/s2 = `(10/5280) * 3600 *3600` ` ` mi/h2

= 24545.454 mi/h2

The time to accelerate to maximum speed and decelerate from maximum speed to zero are same = t1

t1 = `(120 / 24545.454)` h = 0.005 h (rounded to three decimal places, the actual answe is 0.00488888 h)

t1 = 0.005 * 60 minutes = 0.3 minutes.

a) The distance travelled when accelerating  = d1

`d1 = (1/2) * 0.005 h * 120 (mi)/h = 0.3 mi`

Distance travelled during the rest of time = d2

`d2 = 120 (mi/h) * (0.25 - 0.005) h = 29.4 mi`

Therefore total disdance travelled is = 29.4 mi + 0.3 mi = 29.7 mi

 

b)

The distance travelled during acceleration = 0.3 mi

The distnace travelled during deceleration is also same because both are same values.

Time needed to accelerate = time needed to decelerate = 0.005 h

The time travelled in constant speed = (0.25 - 0.005 - 0.005) h

                                                    = 0.24 h

The distance travelled in constant speed = d4

`d4 = 120 (mi/h) * 0.24 h = 28.8 mi`

The total distance travelled = 0.3 mi + 28.8 mi + 0.3 mi = 29.4 mi

 

c)

The minimum time is achieved when train is travelling at maximum speed

T min = `(60 (mi))/(120 (mi)/h) = 0.5 h = 30 minutes`

d)

The distance  = the maximum speed * minimum time

Distance = `120 (mi)/h * (37.5/60) h = 75 mi`

The distance between two stations are 75 miles

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